Question: $f(x, y) = (x + 1)\tan(y)$ What is the partial derivative of $f$ with respect to $x$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(x + 1)\sec^2(y)$ (Choice B) B $\sec^2(y)$ (Choice C) C $\tan(y)$ (Choice D) D $\tan(y) + (x + 1)\sec^2(y)$
Taking a partial derivative with respect to $x$ means treating $y$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial x} &= \dfrac{\partial}{\partial x} \left[ ({x} + 1) \tan(y) \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ {x} \tan(y) \right] + \dfrac{\partial}{\partial x} \left[ \tan(y) \right] \\ \\ &= \tan(y) + 0 \end{aligned}$ Note that $\dfrac{\partial}{\partial x} \left[ \tan(y) \right] = 0$ since we're treating $y$ as a constant. In conclusion, $\dfrac{\partial f}{\partial x} = \tan(y)$